Random Variables and Distributions: Very Difficult Problems with Solutions

Solution:
Since [tex]P(Y=0)=0[/tex], the variable [tex]Z[/tex] is defined with probability [tex]1[/tex] and [tex]Z\gt 0[/tex]; hence [tex]F_Z(z)=0[/tex] for [tex]z\le 0[/tex]. By independence the joint density is [tex]e^{-x}e^{-y}[/tex] for [tex]x,y\gt 0[/tex]. For [tex]z\gt 0[/tex]
[tex]F_Z(z)=P(X\le zY)=\int_0^{\infty}e^{-y}\left(\int_0^{zy}e^{-x}\,dx\right)dy=\int_0^{\infty}e^{-y}\left(1-e^{-zy}\right)dy=1-\frac{1}{1+z}=\frac{z}{1+z}[/tex]
Differentiating, we obtain the density
[tex]f_Z(z)=\begin{cases}\frac{1}{(1+z)^2}, & z\gt 0\\ 0, & z\le 0\end{cases}[/tex]
Check: [tex]\int_0^{\infty}\frac{dz}{(1+z)^2}=1[/tex]. Note that the ratio is not exponential: its tail [tex]P(Z\gt z)=\frac{1}{1+z}[/tex] decays only like [tex]\frac1z[/tex], so [tex]E(Z)=\infty[/tex].

Solution:
We use the formula
[tex]\int_0^{\infty}x^m e^{-x}\,dx=\Gamma(m+1)=m![/tex]
Hence the expected value and the variance of [tex]X[/tex] are
[tex]E(X)=\frac{1}{(n+1)!}\int_0^{\infty}x^{n+2}e^{-x}\,dx=\frac{(n+2)!}{(n+1)!}=n+2[/tex]
[tex]E(X^2)=\frac{(n+3)!}{(n+1)!}=(n+2)(n+3),\qquad D(X)=(n+2)(n+3)-(n+2)^2=n+2[/tex]
Since [tex]2n+5=(n+2)+(n+3)[/tex], the event [tex]|X-(n+2)|\lt n+3[/tex] means [tex]-1\lt X\lt 2n+5[/tex]. Because [tex]X\gt 0[/tex] with probability [tex]1[/tex], this event has the same probability as [tex]X\lt 2n+5[/tex]. Chebyshev's inequality with [tex]\varepsilon=n+3[/tex] gives
[tex]P(X\lt 2n+5)=P\left(|X-(n+2)|\lt n+3\right)\ge 1-\frac{D(X)}{(n+3)^2}=1-\frac{n+2}{(n+3)^2}[/tex]
For example, for [tex]n=1[/tex] the bound is [tex]\frac{13}{16}[/tex], while the exact probability is about [tex]0.97[/tex].

Solution:
Let [tex]U[/tex] and [tex]V[/tex] be the break points. They are independent and uniform on [tex][0, 1][/tex], so the point [tex](U, V)[/tex] has density [tex]1[/tex] in the unit square. By symmetry the cases [tex]U\lt V[/tex] and [tex]U\gt V[/tex] contribute equally (the case [tex]U=V[/tex] has probability [tex]0[/tex]). If [tex]U=u\lt V=v[/tex], the pieces have lengths [tex]u[/tex], [tex]v-u[/tex] and [tex]1-v[/tex]. Hence
[tex]E=2\int_0^1\int_0^v u(v-u)(1-v)\,du\,dv[/tex]
The factor [tex]1-v[/tex] does not depend on [tex]u[/tex], and
[tex]\int_0^v u(v-u)\,du=\frac{v^3}{2}-\frac{v^3}{3}=\frac{v^3}{6}[/tex]
Therefore
[tex]E=2\int_0^1\frac{v^3(1-v)}{6}\,dv=\frac{1}{3}\left(\frac{1}{4}-\frac{1}{5}\right)=\frac{1}{60}[/tex]
Note that the answer is not [tex]\left(\frac{1}{3}\right)^3=\frac{1}{27}[/tex]: each piece has mean length [tex]\frac{1}{3}[/tex], but the three lengths are dependent (their sum is always [tex]1[/tex]), so the expected value of the product is not the product of the expected values.

Solution:
First we find [tex]C[/tex] from the condition [tex]\iint f(x,y)\,dx\,dy=1[/tex]. The integrand is non-negative, so we may integrate in any order. For fixed [tex]y\gt 0[/tex]
[tex]\int_0^{\infty}(y+1)e^{-(y+1)(x+3)}\,dx=e^{-3(y+1)},\qquad \int_0^{\infty}e^{-3(y+1)}\,dy=\frac{e^{-3}}{3}[/tex]
hence [tex]C=3e^3[/tex]. For [tex]x\gt 0[/tex] the marginal density is
[tex]f_X(x)=\int_{-\infty}^{\infty}f(x,y)\,dy=3e^3\int_0^{\infty}(y+1)e^{-(y+1)(x+3)}\,dy[/tex]
We substitute [tex]u=y+1[/tex] and put [tex]a=x+3\gt 0[/tex]; integration by parts gives
[tex]\int_1^{\infty}ue^{-au}\,du=\left[-\frac{u}{a}e^{-au}\right]_1^{\infty}+\frac{1}{a}\int_1^{\infty}e^{-au}\,du=\frac{e^{-a}}{a}+\frac{e^{-a}}{a^2}=\frac{(a+1)e^{-a}}{a^2}[/tex]
Therefore, for [tex]x\gt 0[/tex]
[tex]f_X(x)=3e^3\times \frac{(x+4)e^{-(x+3)}}{(x+3)^2}=\frac{3(x+4)e^{-x}}{(x+3)^2}[/tex]
and [tex]f_X(x)=0[/tex] for [tex]x\le 0[/tex]. Check: since [tex]\frac{3(x+4)e^{-x}}{(x+3)^2}=-\frac{d}{dx}\left(\frac{3e^{-x}}{x+3}\right)[/tex], we get [tex]\int_0^{\infty}f_X(x)\,dx=\frac{3}{3}=1[/tex].

Solution:
For [tex]j=1, \ldots, k[/tex] let [tex]I_j=1[/tex] if the number [tex]j[/tex] is chosen at least once, and [tex]I_j=0[/tex] otherwise. Then the number of distinct numbers is
[tex]S=I_1+I_2+\ldots+I_k[/tex]
In a single choice the number [tex]j[/tex] is missed with probability [tex]1-\frac{1}{k}[/tex], so by the independence of the choices
[tex]P(I_j=0)=\left(1-\frac{1}{k}\right)^n,\qquad E(I_j)=P(I_j=1)=1-\left(1-\frac{1}{k}\right)^n[/tex]
The expected value is linear even though the indicators [tex]I_j[/tex] are dependent, hence
[tex]E(S)=\sum_{j=1}^{k}E(I_j)=k\left[1-\left(1-\frac{1}{k}\right)^n\right][/tex]
Check: for [tex]n=1[/tex] the formula gives [tex]1[/tex], and for fixed [tex]k[/tex] and [tex]n\to\infty[/tex] it tends to [tex]k[/tex].

Solution:
The function [tex]|\cos x|[/tex] is even, and on [tex][0, \pi][/tex] its graph is symmetric about [tex]x=\frac{\pi}{2}[/tex]. From the normalization condition
[tex]1=a\int_{-\pi}^{\pi}|\cos x|\,dx=4a\int_0^{\pi/2}\cos x\,dx=4a\Rightarrow a=\frac{1}{4}[/tex]
The function [tex]\sin x[/tex] is not monotonic on [tex][-\pi, \pi][/tex], so we work with the distribution function [tex]F_Y(y)=P(\sin X\le y)[/tex], [tex]-1\le y\le 1[/tex].
Case [tex]0\le y\le 1[/tex]. Put [tex]\alpha=\arcsin y\in\left[0, \frac{\pi}{2}\right][/tex]. On [tex][-\pi, \pi][/tex] we have [tex]\sin x\le y[/tex] exactly for [tex]x\in[-\pi, \alpha]\cup[\pi-\alpha, \pi][/tex]. By symmetry [tex]P(-\pi\le X\le 0)=\frac{1}{2}[/tex], and
[tex]P(0\le X\le\alpha)=\frac{1}{4}\int_0^{\alpha}\cos x\,dx=\frac{y}{4},\qquad P(\pi-\alpha\le X\le\pi)=\frac{1}{4}\int_{\pi-\alpha}^{\pi}(-\cos x)\,dx=\frac{\sin \alpha}{4}=\frac{y}{4}[/tex]
hence
[tex]F_Y(y)=\frac{1}{2}+\frac{y}{4}+\frac{y}{4}=\frac{1+y}{2}[/tex]
Case [tex]-1\le y\lt 0[/tex]. Put [tex]\beta=-\arcsin y\in\left(0, \frac{\pi}{2}\right][/tex], so [tex]\sin \beta=-y[/tex]. Then [tex]\sin x\le y[/tex] exactly for [tex]x\in[-\pi+\beta, -\beta][/tex], and since [tex]f[/tex] is even,
[tex]F_Y(y)=P(\beta\le X\le\pi-\beta)=\frac{1}{4}\left(\int_{\beta}^{\pi/2}\cos x\,dx+\int_{\pi/2}^{\pi-\beta}(-\cos x)\,dx\right)=\frac{2(1-\sin \beta)}{4}=\frac{1+y}{2}[/tex]
Thus [tex]F_Y(y)=\frac{1+y}{2}[/tex] on [tex][-1, 1][/tex] (with [tex]F_Y(y)=0[/tex] for [tex]y\lt -1[/tex] and [tex]F_Y(y)=1[/tex] for [tex]y\gt 1[/tex]), i.e. [tex]Y[/tex] is uniformly distributed on [tex](-1, 1)[/tex]. Differentiating, we get
[tex]f_Y(y)=\begin{cases}\frac{1}{2}, & |y|\lt 1\\ 0, & |y|\ge 1\end{cases}[/tex]
The same result follows from the formula for non-monotonic transformations: each [tex]y\in(-1, 0)\cup(0, 1)[/tex] has two preimages [tex]x_1, x_2[/tex] in [tex][-\pi, \pi][/tex], and [tex]f_Y(y)=\frac{f(x_1)}{|\cos x_1|}+\frac{f(x_2)}{|\cos x_2|}=2\times \frac{1}{4}=\frac{1}{2}[/tex].

Solution:
Write [tex]\gamma=\gamma_1+\gamma_2[/tex], where [tex]\gamma_1[/tex] is the number of white balls coming from the first box and [tex]\gamma_2[/tex] the number coming from the second box; they are independent. [tex]\gamma_1[/tex] is binomial with [tex]n=2[/tex], [tex]p=\frac{1}{2}[/tex], and [tex]\gamma_2[/tex] is hypergeometric:
[tex]P(\gamma_1=0)=\frac{1}{4},\quad P(\gamma_1=1)=\frac{1}{2},\quad P(\gamma_1=2)=\frac{1}{4}[/tex]
[tex]P(\gamma_2=0)=\frac{\binom{2}{2}}{\binom{6}{2}}=\frac{1}{15},\quad P(\gamma_2=1)=\frac{4\times 2}{15}=\frac{8}{15},\quad P(\gamma_2=2)=\frac{\binom{4}{2}}{\binom{6}{2}}=\frac{6}{15}[/tex]
Combining them, we obtain the distribution of [tex]\gamma[/tex]:
[tex]P(\gamma=0)=\frac{1}{60},\ P(\gamma=1)=\frac{10}{60},\ P(\gamma=2)=\frac{23}{60},\ P(\gamma=3)=\frac{20}{60},\ P(\gamma=4)=\frac{6}{60}[/tex]
The third box contains [tex]4[/tex] balls, so [tex]P(A\mid\gamma=k)=\frac{k}{4}[/tex]. By the total probability formula
[tex]P(A)=\sum_{k}\frac{k}{4}P(\gamma=k)=\frac{E(\gamma)}{4},\qquad E(\gamma)=\frac{0+10+46+60+24}{60}=\frac{7}{3},\qquad P(A)=\frac{7}{12}[/tex]
By Bayes' formula [tex]P(\gamma=k\mid A)=\frac{\frac{k}{4}P(\gamma=k)}{P(A)}[/tex], hence
[tex]E(\gamma\mid A)=\sum_{k}k\,P(\gamma=k\mid A)=\frac{E(\gamma^2)}{4P(A)}=\frac{E(\gamma^2)}{E(\gamma)}[/tex]
Here
[tex]E(\gamma^2)=\frac{0+10+92+180+96}{60}=\frac{63}{10}\Rightarrow E(\gamma\mid A)=\frac{63}{10}\times \frac{3}{7}=\frac{27}{10}[/tex]
This is larger than [tex]E(\gamma)=\frac{7}{3}[/tex], as expected: drawing a white ball makes the compositions with many white balls more likely.

Solution:
The roots are real if and only if [tex]U_2^2\ge 4U_1U_3[/tex] (the event [tex]U_1=0[/tex] has probability [tex]0[/tex]). Fix [tex]U_2=b[/tex] and put [tex]\lambda=\frac{b^2}{4}\in\left[0, \frac{1}{4}\right][/tex]. The point [tex](a, c)=(U_1, U_3)[/tex] is uniformly distributed in the unit square, so [tex]P(ac\le\lambda)[/tex] equals the area of the part of the square where [tex]ac\le\lambda[/tex]. For [tex]a\le\lambda[/tex] every [tex]c\in[0, 1][/tex] works, and for [tex]a\gt\lambda[/tex] we need [tex]c\le\frac{\lambda}{a}\lt 1[/tex]. Hence
[tex]P(ac\le\lambda)=\lambda+\int_{\lambda}^{1}\frac{\lambda}{a}\,da=\lambda(1-\ln\lambda)[/tex]
By the total probability formula over [tex]U_2[/tex] (density [tex]1[/tex] on [tex][0, 1][/tex])
[tex]P=\int_0^1\frac{b^2}{4}\left(1-\ln\frac{b^2}{4}\right)db=\frac{1}{12}-\frac{1}{4}\int_0^1 b^2(2\ln b-\ln 4)\,db[/tex]
Integration by parts gives
[tex]\int_0^1 b^2\ln b\,db=\left[\frac{b^3}{3}\ln b\right]_0^1-\int_0^1\frac{b^2}{3}\,db=-\frac{1}{9},\qquad \int_0^1 b^2\,db=\frac{1}{3}[/tex]
Therefore
[tex]P=\frac{1}{12}-\frac{1}{4}\left(-\frac{2}{9}-\frac{2\ln 2}{3}\right)=\frac{1}{12}+\frac{1}{18}+\frac{\ln 2}{6}=\frac{5+6\ln 2}{36}\approx 0.254[/tex]